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(C) Given: Battery voltage $E = 4.5 \ V$,Barrier potential $V_b = 0.5 \ V$,Resistance $R = 100 \ \Omega$.The effective voltage $V$ across the resistor

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$40$

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(C) Given: Battery voltage $E = 4.5 \ V$,Barrier potential $V_b = 0.5 \ V$,Resistance $R = 100 \ \Omega$.The effective voltage $V$ across the resistor is the battery voltage minus the barrier potential of the diode:$V = E - V_b = 4.5 \ V - 0.5 \ V = 4.0 \ V$.Using Ohm's law,the current $I$ in the circuit is:$I = \frac{V}{R} = \frac{4.0 \ V}{100 \ \Omega} = 0.04 \ A$.To convert the current to milliamperes $(mA)$:$I = 0.04 \ 	ext{A} 	imes 1000 \ 	ext{mA/A} = 40 \ mA$.

As shown in the figure,a diode is connected to an external resistor. Assume that the barrier potential developed in the diode is $0.5 \ V$. Find the value of the current in the circuit in milliamperes. (in $mA$)

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